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What is the Lipschitz constant of $e^{-x}$ if it has one.

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2 Answers 2

If a function $f$ is differentiable and has a bounded first derivative, then it is Lipschitz continuous with a Lipschitz constant of $K = \sup | f'(x) |$.

For the function $f(x) = e^{-x}$ on $x \in (0, \infty)$, $K = 1$.

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It is indeed Lipschitz with a Lipschitz constant of $1$. In order to show that this is the case, one may use the mean value theorem.

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